Answer:
a.
[tex]x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )[/tex]
b. x = -8 and y = 4
Step-by-step explanation:
This question is incomplete. I will type the complete question below before giving my solution.
For integers a, b, c, consider the linear Diophantine equation
[tex]ax+by=c[/tex]
Suppose integers x0 and yo satisfy the equation; that is,
[tex]ax_0+by_0 = c[/tex]
what other values
[tex]x = x_0+h[/tex] and [tex]y=y_0+k[/tex]
also satisfy ax + by = c? Formulate a conjecture that answers this question.
Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.
Can you find other integers x and y such that 6x + 15y = 12?
How many other pairs of integers x and y can you find ?
Can you find infinitely many other solutions?
From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that
[tex]as+bt=gcd(a,b)[/tex]
the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.
Multiplying as + bt = gcd(a,b) through by k you get
[tex]a(sk) + b(tk) = gcd(a,b)k = c[/tex]
So this gives one solution, with x = sk and y = tk.
Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get
[tex]a(x_1-x) + b(y_1-y)=0[/tex]
Therefore,
[tex]a(x_1-x) = b(y-y_1)[/tex]
This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,
[tex]y = y_1+r(\frac{a}{gcd(a, b)})[/tex] for some integer r. Substituting into the equation
[tex]a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba[/tex]
or
[tex]x = x_1-r(\frac{b}{gcd(a, b)} )[/tex]
Thus if ax1 + by1 = c is any solution, then all solutions are of the form
[tex]x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )[/tex]
In order to find all integer solutions to 6x + 15y = 12
we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.
[tex]15 = 6*2+3\\6=3*2+0[/tex]
Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.
We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,
[tex]3 = 15-6*2[/tex]
Because 4 multiplies 3 to give 12, we multiply by 4
[tex]12 = 15*4-6*8[/tex]
So one solution is
[tex]x=-8[/tex] & [tex]y = 4[/tex]
All other solutions will have the form
[tex]x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r[/tex]
where [tex]r[/tex] ∈ Ζ
Hence by putting r values, we get many (x, y)
